Question:

Step 1: Understanding what "into function" means For $f$ to be an into function (not onto), the range of $f$ must be a proper subset of the codomain $\mathbb{R}$. This means there must exist at least one element in $\mathbb{R}$ that is not in the range of $f$. Step 2: Finding the range for $x \geq 2$ For the piece $f(x) = 2x + \alpha^2$ when $x \geq 2$: Since $x \geq 2$, we have: $$f(x) = 2x + \alpha^2 \geq 2(2) + \alpha^2 = 4 + \alpha^2$$ As $x$ increases without bound, $f(x)$ also increases without bound. Therefore, the range for this piece is: $$[4 + \alpha^2, \infty)$$ Step 3: Finding the range for $x < 2$ For the piece $f(x) = \frac{\alpha x}{2} + 10$ when $x < 2$, we consider the sign of $\alpha$: **Case 1: If $\alpha > 0$** - As $x \to -\infty$, we have $f(x) \to -\infty$ - As $x \to 2^-$, we have $f(x) \to \frac{\alpha(2)}{2} + 10 = \alpha + 10$ - The range for this piece is $(-\infty, \alpha + 10)$ **Case 2: If $\alpha < 0$** - As $x \to -\infty$, we have $f(x) \to +\infty$ - As $x \to 2^-$, we have $f(x) \to \alpha + 10$ - The range for this piece is $(\alpha + 10, \infty)$ **Case 3: If $\alpha = 0$** - $f(x) = 10$ for all $x < 2$ - The range for this piece is $\{10\}$ Step 4: Determining the condition for $f$ to be into For $\alpha > 0$ (positive case), the combined range of $f$ is: $$(-\infty, \alpha + 10) \cup [4 + \alpha^2, \infty)$$ For $f$ to be an into function, there must be a gap between these two intervals. This requires: $$\alpha + 10 < 4 + \alpha^2$$ Rearranging: $$\alpha^2 - \alpha - 6 > 0$$ Factoring: $$(\alpha - 3)(\alpha + 2) > 0$$ For positive values of $\alpha$, this inequality is satisfied when: $$\alpha > 3$$ Step 5: Finding the least integral positive value The least integral positive value of $\alpha$ that satisfies $\alpha > 3$ is: $$\alpha = 4$$ We can verify: when $\alpha = 4$, the range is $(-\infty, 14) \cup [20, \infty)$, which leaves the interval $[14, 20)$ uncovered in $\mathbb{R}$, confirming that $f$ is indeed an into function. **Final Answer: The least integral positive value of $\alpha$ is $\boxed{4}$, which corresponds to Option 4.**